Question Number 198282 by MathedUp last updated on 16/Oct/23
Answered by witcher3 last updated on 25/Oct/23
![{ ((2cos(t))),((2sin(t))) :},t∈[0,2π] circle radius=2 origine(0,0) ∫_C (−((xy)/5)dx+2ydy)=∫∫_D (∂((2y)/∂x)−(∂/∂y)(−((xy)/5)))dA =∫∫_D ((x/5))dA D disc of radius 2 ,center(0,0) x=rcos(a),dA=rdrda =∫_0 ^(2π) ∫_0 ^2 ((1/5)rcos(a))rdrda =∫_0 ^2 (r^2 /5).∫_0 ^(2π) cos(a)da=0](https://www.tinkutara.com/question/Q198885.png)
$$\begin{cases}{\mathrm{2cos}\left(\mathrm{t}\right)}\\{\mathrm{2sin}\left(\mathrm{t}\right)}\end{cases},\mathrm{t}\in\left[\mathrm{0},\mathrm{2}\pi\right]\:\mathrm{circle}\:\mathrm{radius}=\mathrm{2}\:\mathrm{origine}\left(\mathrm{0},\mathrm{0}\right) \\ $$$$\int_{\mathrm{C}} \left(−\frac{\mathrm{xy}}{\mathrm{5}}\mathrm{dx}+\mathrm{2ydy}\right)=\int\int_{\mathrm{D}} \left(\partial\frac{\mathrm{2y}}{\partial\mathrm{x}}−\frac{\partial}{\partial\mathrm{y}}\left(−\frac{\mathrm{xy}}{\mathrm{5}}\right)\right)\mathrm{dA} \\ $$$$=\int\int_{\mathrm{D}} \left(\frac{\mathrm{x}}{\mathrm{5}}\right)\mathrm{dA} \\ $$$$\mathrm{D}\:\mathrm{disc}\:\mathrm{of}\:\mathrm{radius}\:\mathrm{2}\:,\mathrm{center}\left(\mathrm{0},\mathrm{0}\right) \\ $$$$\mathrm{x}=\mathrm{rcos}\left(\mathrm{a}\right),\mathrm{dA}=\mathrm{rdrda} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \int_{\mathrm{0}} ^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{5}}\mathrm{rcos}\left(\mathrm{a}\right)\right)\mathrm{rdrda} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{r}^{\mathrm{2}} }{\mathrm{5}}.\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{cos}\left(\mathrm{a}\right)\mathrm{da}=\mathrm{0} \\ $$$$ \\ $$